(Theorem REF )
Let \(A\)  be a simple unital separable exact C\(^*\) -algebra which is approximately divisible and of real rank zero, and let \(a,b \in A\)  be positive with \([a]=[b]\)  in \(\mathrm {Cu}(A)\) .
Since \(A\)  is \(\mathcal {Z}\) -stable by Theorem 2.3 of [1]}, it is either stably finite (when it has a trace) or purely infinite (when it is traceless) ([2]}). In the purely infinite case, the Cuntz classes of any two nonzero positive elements are the same, so the line segment \((1-t)a+tb, \ t \in [0,1]\) , defines a homotopy between \(a\)  and \(b\) . This segment consists entirely of nonzero positive elements and the Theorem follows. We may thus assume that \(A\)  is stably finite, whence \(\mathrm {T}(A) \ne \emptyset \) . In fact, \(A\)  has stable rank one by Theorem 6.7 of [2]}.
If \([a]=[b]\)  is compact in \(\mathrm {Cu}(A)\)  then zero is not an accumulation point of the spectrum of \(a\)  or \(b\)  by Proposition 2.23 of [4]}. By Lemma REF  we may assume that \(a\)  and \(b\)  are projections and the desired result follows from Proposition REF .
It remains to address the case where \([a]=[b]\)  is not compact, so that \(a\)  and \(b\)  have zero as an accumulation point of their spectrum. By Proposition REF , we may assume that there are sequences of projections \(p_1,p_2,\ldots \)  and \(q_1,q_2,\ldots \)  in \(\overline{aAa}\)  and \(\overline{bAb}\) , respectively, with the following properties:

\(p_ip_j = 0 = q_iq_j\)  for every pair \(i \ne j\) ;

\(p_i \sim q_i\)  for each \(i \in \mathbb {N}\) ;


\(a = \bigoplus _{i=1}^\infty \frac{1}{2^i} p_i \ \mathrm {and} \ b = \bigoplus _{i=1}^\infty \frac{1}{2^i} q_i.\) 

For convenience we set \(s_k = \oplus _{i=1}^k 1/2^i\) . Assume that we have found a continuous path \(u_t:[0,s_k] \rightarrow \mathcal {U}(A)\)  with the property that \(u_0 = 1_A\)  and
\(u_{s_k} \left( \oplus _{i=1}^k p_i \right) u_{s_k}^* = \oplus _{i=1}^k q_i.\) 
(The existence of this path for \(k=1\)  follows from (ii) above and Lemma REF .) Let us show that \(u_t\)  can be extended to \([0,s_{k+1}]\)  so that (REF ) holds with \(k+1\)  in place of \(k\)  and moreover that
\(u_t \left( \oplus _{i=1}^k p_i \right) u_t^* = \oplus _{i=1}^k q_i, \ \forall t \in [s_k,s_{k+1}].\) 
Set
\(Q_k = 1_A - \left( \oplus _{i=1}^k q_i \right).\) 
Note that \(Q_k A Q_k\)  is approximately divisible by [5]}[Corollary 2.9], and that \(u_{s_k} p_{k+1} u_{s_k}^* \sim q_{k+1}\)  in \(Q_k A Q_k\) . It follows from Lemma REF  that there is a continuous path
\(\tilde{v}_t:[s_k,s_{k+1}] \rightarrow \mathcal {U}(Q_k A Q_k)\) 
such that \(\tilde{v}_{s_k} = Q_k\)  and
\(\tilde{v}_{s_{k+1}} (u_{s_k} p_{k+1} u_{s_k}^*) \tilde{v}_{s_{k+1}}^* = q_{k+1}.\) 
Set \(v_t = \tilde{v}_t + (1_A-Q_k), \ t \in [s_k,s_{k+1}]\) , so that \(v_t \in \mathcal {U}(A)\)  and \(v_{s_k} = 1_A\) . We now define our extension of \(u_t:[0,s_{k+1}] \rightarrow \mathcal {U}(A)\)  via
\(u_t = \left\lbrace  \begin{array}{rcl} u_t & = & t \in [0,s_k) \\ v_t u_{s_k} & = & t \in [s_k,s_{k+1}] \end{array} \right. .\) 
This extension is continuous since \(v_{s_k} = 1_A\) . Also,
\(u_{s_{k+1}} \left( \oplus _{i=1}^{k+1} p_i \right) u_{s_{k+1}}^* & = v_{s_{k+1}} \left( u_{s_{k}} \left( \oplus _{i=1}^{k+1} p_i \right) u_{s_{k}}^* \right) v_{s_{k+1}}^* \\& = v_{s_{k+1}} \left( u_{s_{k}} \left( \oplus _{i=1}^{k} p_i \right) u_{s_{k}}^* \right) v_{s_{k+1}}^* \\& \hspace{28.45274pt} + \ v_{s_{k+1}} u_{s_k} p_{i+1} u_{s_{k}}^* v_{s_{k+1}}^* \\& = v_{s_{k+1}} \left( \oplus _{i=1}^k q_i \right) v_{s_{k+1}}^* \\& \hspace{28.45274pt} + \ \ \tilde{v}_{s_{k+1}} u_{s_k} p_{i+1} u_{s_{k}}^* \tilde{v}_{s_{k+1}}^* \\&= \left( \oplus _{i=1}^k q_i \right) + q_{k+1} \\&= \oplus _{i=1}^{k+1} q_i\) 

satisfying (REF ) with \(k+1\)  in place of \(k\) , as required. Note also that for \(t \in [s_k,s_{k+1}]\)  we have
\(u_t \left( \oplus _{i=1}^k p_i \right) u_t^* & = v_t \left( u_{s_{k}} \left( \oplus _{i=1}^k p_i \right) u_{s_{k}}^* \right) v_t^* \\& = v_t \left( \oplus _{i=1}^{k} q_i \right) v_t^* \\&= \oplus _{i=1}^k q_i\) 

as required by (REF ). This completes our extension of \(u_t\)  to \([0,s_{k+1}]\) . Beginning at \(t=0\)  and continuing these extensions inductively yields a continuous path \(u_t:[0,1) \rightarrow \mathcal {U}(A)\)  such that
\(u_t \left( \oplus _{i=1}^l p_i \right) u_t^* = \oplus _{i=1}^l q_i, \ \forall t \ge s_k, \ \forall l \le k.\) 
Define \(h:[0,1] \rightarrow A\)  as follows:
\(h(t) = \left\lbrace  \begin{array}{rl} u_t a u_t^*, & t \in [0,1) \\ b, &t = 1 \end{array} \right.\) 
Clearly, \([h(t)]=[a]\)  in \(\mathrm {Cu}(A)\)  for every \(t\)  in \([0,1]\) , and \(h(t)\)  is continuous at every \(t\)  in \([0,1)\)  since \(u_t\)  is. It remains to prove that \(h(t)\)  is continuous at \(t=1\) . Recall that \(s_k \rightarrow 1\)  as \(k \rightarrow \infty \) . For \(t \ge s_k\)  we have
\(\Vert  h(1)-h(t) \Vert  &= \Vert  b - u_t a u_t* \Vert  \\&=\left\Vert  \bigoplus _{i=1}^\infty \frac{1}{2^i} q_i - u_t \left( \bigoplus _{i=1}^\infty \frac{1}{2^i} p_i \right) u_t^* \right\Vert  \\& \stackrel{(\ref {conjequal2})}{=} \left\Vert  \bigoplus _{i=k+1}^\infty \frac{1}{2^i} q_i - u_t \left( \bigoplus _{i=k+1}^\infty \frac{1}{2^i} p_i \right) u_t^* \right\Vert  \\& \le \left\Vert  \bigoplus _{i=k+1}^\infty \frac{1}{2^i} q_i \right\Vert  + \left\Vert  \bigoplus _{i=k+1}^\infty \frac{1}{2^i} p_i \right\Vert  \\&= \frac{1}{2^{k+1}} + \frac{1}{2^{k+1}} \\&= \frac{1}{2^k}\) 

This shows that \(h(t)\)  is continuous at \(t=1\) , completing the proof.
